Area And Perimeter Of Different Types Of Triangle

Perimeter

The perimeter is the length of the boundary around any closed figure. Units of perimeter's measurement are taken the same as the units of length.

The few examples of units of perimeter are meter (m), centimeter (cm) etc..

Triangle

The perimeter of a triangle is the total length of the boundary that consists of its three sides. Therefore, the sum of the length of three sides of a triangle is called its perimeter.

Perimeter of triangle ΔABC

In ΔABC, AB = a, BC = b, and CA = c

So, perimeter of ΔABC is sum of length of sides AB, BC and CA

i.e. perimeter of ΔABC = a + b + c

Equilateral triangle

An equilateral triangle has all of its sides of equal length.

∴ Perimeter of an equilateral triangle can be calculated by adding up its all three sides or by multiplying length of its side with 3.

Equilateral triangle ΔABC with sides length a

In equilateral triangle ΔABC, AB = a, BC = a and CA = a

So, perimeter of equilateral ΔABC = a + a + a = 3a

Isosceles triangle

In an isosceles triangle, any two sides are of equal length.

Isosceles triangle ΔABC with sides length a, b and a

AB = a, BC = a and CA = b

So, perimeter of isosceles ΔABC = a + a + b = 2a + b

Area

Area is the total amount of space occupied by any closed figure. Area is measured in square units.

The few examples of units of area are meter² or m² read as meter square, centimeter² or cm² read as centimeter square etc.

Area of triangle

Area of any triangle = $\frac{1}{2}\times \mathrm{base}\times \mathrm{height}$

Area of triangle ΔABC

In ΔABC, BC is base, length of BC = b

and AO is height, length of AO = h

Therefore, area of ΔABC = $\frac{1}{2}\times b\times h$

Area of equilateral triangle

In equilateral, where all sides of triangle are equal in length, its area is calculated as:

Area of ΔABC = $\frac{\sqrt{3}}{4}{\mathrm{a}}^2$ where a is the length of side

Here, in right angle ΔAOC

$a^2=h^2+{\left(\frac{a}{2}\right)}^2$       by Pythagoras theorem

$a^2=h^2+\frac{a^2}{4}$

$a^2-\frac{a^2}{4}=h^2$

$\frac{4a^2-a^2}{4}=h^2$

$\frac{3a^2}{4}=h^2$

$\frac{\sqrt{3}a}{2}=h$

Area of equilateral Δ ABC = $\frac{1}{2}\times a\times \frac{\sqrt{3}a}{2}$

= $\frac{\sqrt{3}}{4}{\mathrm{a}}^2$

Area of isosceles triangle

area of isosceles ΔABC = $\frac{b\sqrt{{\mathrm{4a}}^2-b^2}}{4}$

Here, in right angle ΔAOC

$a^2=h^2+{\left(\frac{b}{2}\right)}^2$           by Pythagoras theorem

$a^2=h^2+\frac{b^2}{4}$

$a^2-\frac{b^2}{4}=h^2$

$\frac{{\mathrm{4a}}^2-b^2}{4}=h$

$\frac{\sqrt{{\mathrm{4a}}^2-b^2}}{2}=h$

we know, area of Δ = $\frac{1}{2}\times b\times h$

∴ area of isosceles ΔABC = $\frac{1}{2}\times b\times \frac{\sqrt{{\mathrm{4a}}^2-b^2}}{2}$

= $\frac{b\sqrt{{\mathrm{4a}}^2-b^2}}{4}$